The question I’d like to explore in this post is how Ampere’s law, the relationship between the line integral of the magnetic field to current (i.e. the enclosed current)
\begin{equation}\label{eqn:flux:20}
\oint_{\partial A} d\Bx \cdot \BH = -\int_A \ncap \cdot \BJ,
\end{equation}
generalizes to geometric algebra where Maxwell’s equations for a statics configuration (all time derivatives zero) is
\begin{equation}\label{eqn:flux:40}
\spacegrad F = J,
\end{equation}
where the multivector fields and currents are
\begin{equation}\label{eqn:flux:60}
\begin{aligned}
F &= \BE + I \eta \BH \\
J &= \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_\txtm – \BM }.
\end{aligned}
\end{equation}
Here (fictitious) the magnetic charge and current densities that can be useful in antenna theory have been included in the multivector current for generality.

My presumption is that it should be possible to utilize the fundamental theorem of geometric calculus for expressing the integral over an oriented surface to its boundary, but applied directly to Maxwell’s equation. That integral theorem has the form
\begin{equation}\label{eqn:flux:80}
\int_A d^2 \Bx \boldpartial F = \oint_{\partial A} d\Bx F,
\end{equation}
where \( d^2 \Bx = d\Ba \wedge d\Bb \) is a two parameter bivector valued surface, and \( \boldpartial \) is vector derivative, the projection of the gradient onto the tangent space. I won’t try to explain all of geometric calculus here, and refer the interested reader to [1], which is an excellent reference on geometric calculus and integration theory.

This is not nearly as nice as the magnetic flux relationship which was nicely split with the current and fields nicely separated. The \( d\Bx F \) product has all possible grades, as does the \( d^2 \Bx J \) product (in general). Observe however, that the normal term on the right has only grades 1,2, so we can split our line integral relations into pairs with and without grade 1,2 components
\begin{equation}\label{eqn:flux:140}
\begin{aligned}
\oint_{\partial A} \gpgrade{d\Bx F}{0,3}
&=
\int_A dA \gpgrade{ I \ncap J }{0,3} \\
\oint_{\partial A} \gpgrade{d\Bx F}{1,2}
&=
\int_A dA \lr{ \gpgrade{ I \ncap J }{1,2} – \lr{ \ncap \cdot \spacegrad } I F }.
\end{aligned}
\end{equation}

where \( \BB \) is an applied external magnetic field and \( \Bm \) is the magnetic dipole for the current in question. These results (and a similar one derived earlier for the vector potential \( \BA \)) all follow from
an analysis of localized current densities \( \BJ \), evaluated far enough away from the current sources.

For the force and torque, the starting point for the force is one that had me puzzled a bit. Namely

For the point particle, this is the force on the particle when it is in the external field \( BB \). i.e. this is the force at the position of the particle. My question is what does it mean to sum all the forces on the charge distribution over all space.
How can a force be applied over all, as opposed to a force applied at a single point, or against a surface?

In the special case of a localized current density, this makes some sense. Considering the other half of the force equation \( \BF = \ddt{}\int \rho_m \Bv dV \), where \( \rho_m \) here is mass density of the charged particles making up the continuous current distribution. The other half of this \( \BF = m\Ba \) equation is also an average phenomena, so we have an average of sorts on both the field contribution to the force equation and the mass contribution to the force equation. There is probably a centre-of-mass and centre-of-current density interpretation that would make a bit more sense of this continuum force description.

It’s kind of funny how you can work through all the detailed mathematical steps in a book like Jackson, but then go right back to the beginning and say “Hey, what does that even mean”?

Force

Moving on from the pondering of the meaning of the equation being manipulated, let’s do the easy part, the derivation of the results that Jackson comes up with.

Writing out \ref{eqn:magnetostaticsJacksonNotesForceAndTorque:80} in coordinates

This shows that the scalar part of the equation is zero, provided the normal component of \( \BJ/\Abs{\Bx – \Bx’} \) vanishes on the boundary of the infinite sphere. This leaves the Biot-Savart law as a bivector equation

See fig. 1 for a sketch of energy vs momentum. The asymptotes are the limiting cases when \( m c^2 \rightarrow 0 \). The \( + \) branch is what we usually associate with particles. What about the other energy states. For Fermions Dirac argued that the lower energy states could be thought of as “filled up”, using the Pauli principle to leave only the positive energy states available. This was called the “Dirac Sea”. This isn’t a good solution, and won’t work for example for Bosons.

fig. 1. Dirac equation solution space

Another way to rationalize this is to employ ideas from solid state theory. For example consider a semiconductor with a valence and conduction band as sketched in fig. 2.

fig. 2. Solid state valence and conduction band transition

A photon can excite an electron from the valence band to the conduction band, leaving all the valence band states filled except for one (a hole). For an electron we can use almost the same picture, as sketched in fig. 3.

fig. 3. Pair creation

A photon with energy \( E_k – (-E_k) \) can create a positron-electron pair from the vacuum, where the energy of the electron and positron pair is \( E_k \).

At high enough energies, we can see this pair creation occur.

Zitterbewegung

If a particle is created at a non-eigenstate such as one on the asymptotes, then oscillations between the positive and negative branches are possible as sketched in fig. 4.

fig. 4. Zitterbewegung oscillation

Only “vertical” oscillations between the positive and negative locations on these branches is possible since those are the points that match the particle momentum. Examining this will be the aim of one of the problem set problems.

Here the speed of light \( c \) is the slope of the line in the plots above. We can think of this current density as right movers minus the left movers. Any state that is given can be thought of as a combination of right moving and left moving states, neither of which are eigenstates of the free particle Hamiltonian.

Potential step

The next logical thing to think about, as in non-relativistic quantum mechanics, is to think about what occurs when the particle hits a potential step, as in fig. 5.

fig. 5. Reflection off a potential barrier

The approach is the same. We write down the wave functions for the \( V = 0 \) region (I), and the higher potential region (II).

The eigenstates are found on the solid lines above the asymptotes on the right hand movers side as sketched in fig. 6. The right and left moving designations are based on the phase velocity \( \PDi{k}{E} \) (approaching \( \pm c \) on the top-right and top-left quadrants respectively).